 | Thomas Aloysius O'Donahue - Mine surveying - 1911 - 288 pages
...method of casting out areas is that known as " equalising," which is based on Euclid's proposition of " triangles on the same base and between the same parallels are equal." The method consists in constructing a triangle equal to the figure, the area of which is required,... | |
 | Joseph Harrison, George Albert Baxandall - Geometry, Descriptive - 1913 - 714 pages
...DK. Then DKG is a triangle having an area equal to the polygon. This solution is based on the theorem that triangles on the same base and between the same parallels are equal in area. 31. PROBLEM. — To construct a square equal in area to a given rectangle. Determine a mean proportional... | |
 | Newfoundland Council of Higher Education - 1914 - 228 pages
...along AD and the point B comes to K. Prove that CH\s equal to DK. (8) 6. Assuming that parallelograms, on the same base and between the same parallels, are equal in area, prove that the area of a triangle can be found from the formula : Area = ^ base X height. A point 0... | |
 | United States. Office of Education - Agricultural colleges - 1917 - 1336 pages
...in F, and F is joined to B. Prove that the angles BFA, EFA are equal. 3. Prove that parallelograms on the same base and between the same parallels are equal in area. ABCD, AEFG are two parallelograms having a common point at A, and having the vertex E on BC, and the vertex... | |
 | Alexander Caswell Ellis - Education - 1917 - 1098 pages
...in F, and F is joined to B. Prove that the angles BFA, EFA are equal. 3. Prove that parallelograms on the same base and between the same parallels are equal in area. ABCD, AEFG are two parallelograms having a common point at A, and having the vertex E on BC, and the vertex... | |
 | Raymond Clare Archibald - Mathematics - 1918 - 310 pages
...in F, and F is joined to B. Prove that the angles BFA, EFA are equal. 3. Prove that parallelograms on the same base and between the same parallels are equal in area. ABCD, AEFG are two parallelograms having a common point at A, and having the vertex E on BC, and the vertex... | |
 | United States. Office of Education - 1918 - 1128 pages
...in F, and F is joined to B. Prove that the angles BFA, EFA are equal. 3. Prove that parallelograms on the same base and between the same parallels are equal in area. ABCD, AEFG are two parallelograms having a common point at A, and having the vertex E on BC, and the vertex... | |
 | John William Angles - Measurement - 1919 - 200 pages
...lines. .-. Area of triangle = | x base x perpendicular height = \bh <s— — —— — • —SB^ Triangles on the same base and between the same parallels are equal in area. The line joining the apex of a triangle to the middle point of the opposite side is called a median.... | |
 | Bennett Hooper Brough - Mine surveying - 1920 - 510 pages
...line parallel to AD, meeting CD produced, at G. Join A G. The method depends upon Euclids' theorem that triangles on the same base and between the same parallels are equal to one another. By using a parallel ruler and a pricker the drawing in of all the constructional lines,... | |
 | Newfoundland Council of Higher Education - 1922 - 204 pages
...Grade.) Wednesday, June. 28th, 1922.— Morning, 9 to 12. Recognized abbreviations may be used. 1. Prove that triangles on the same base and between the same parallels are equal in area. A straight line parallel to the diagonal DB of a parallelogram A BCD meets CD and CB in E and F respectively.... | |
| |
Prove that parallelograms on the same base and between the same parallels are equal in area. 
